Hi,
I'd like to process files in a directory which is in fact a short cut link
to another directory (under windows XP).
If the path to the directory is for instance called c:\test. I have tried
both following code snipets for printing all names of files in the
directory:
++ snippet 1++
for filename in glob.glob( os.path.join(r'c:\test', '*') ):
print filename
++ snippet 2++
for filename in glob.glob( os.path.join(r'c:\test.lnk', '*') ):
print filename
Neither solution works.
Thanks very much for any help.
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